Problem: Rewrite the equation by completing the square. $2 x^{2} +7 x +6 = 0$ $(x + $
Explanation: $\begin{aligned} 2 x^2 +7 x +6&=0 \\\\ 2 x^2 +7 x &=-6 \\\\ x^2 +\dfrac{7}{2} x&=-3 \end{aligned}$ Now we want to complete $x^2 +\dfrac{7}{2} x$ into a perfect square. To do that, we should add $\left(\dfrac{{\frac{7}{2}}}{2}\right)^2={\dfrac{49}{16}}$ to it: $x^2{+\dfrac{7}{2}}x + {\dfrac{49}{16}}=\left(x +\dfrac{7}{4} \right)^2$ $\begin{aligned} x^2 +\dfrac{7}{2} x&=-3 \\\\ x^2 +\dfrac{7}{2} x + {\dfrac{49}{16}}&=-3 + {\dfrac{49}{16}} \\\\ \left(x +\dfrac{7}{4} \right)^2&=\dfrac{1}{16} \end{aligned}$ In conclusion, the equation after completing the square is written as: $\left(x +\dfrac{7}{4} \right)^2=\dfrac{1}{16}$